How do you use implicit differentiation to find dy/dx given #x^2-2y^2+x+3y-4=0#?

Question

Emily Ammerman · Accepted Answer

#dy/dx = (2x+1)/(2y-3)# Differentiate both sides of the equation with respect to #x#: #d/dx (x^2-2y^2+x+3y-4) = 0# #2x -2ydy/dx+1 +3dy/dx = 0# #2x +1 = (2y-3) dy/dx# #dy/dx = (2x+1)/(2y-3)#

Emily Ammerman · Answer

undefined To find $ \frac{dy}{dx} $ using implicit differentiation for the equation $ x^2 - 2y^2 + x + 3y - 4 = 0 $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Treat $ y $ as a function of $ x $ and use the chain rule where necessary.
3. Solve the resulting equation for $ \frac{dy}{dx} $.

Here's a step-by-step breakdown:

1. $ \frac{d}{dx}(x^2 - 2y^2 + x + 3y - 4) = \frac{d}{dx}(0) $
2. Apply the derivative rules:
   - $ \frac{d}{dx}(x^2) = 2x $
   - $ \frac{d}{dx}(-2y^2) = -4y \frac{dy}{dx} $ (using the chain rule)
   - $ \frac{d}{dx}(x) = 1 $
   - $ \frac{d}{dx}(3y) = 3 \frac{dy}{dx} $ (using the chain rule)
   - $ \frac{d}{dx}(-4) = 0 $
3. Plug these derivatives into the equation and solve for $ \frac{dy}{dx} $:
   $ 2x - 4y \frac{dy}{dx} + 1 + 3 \frac{dy}{dx} = 0 $
   $ (2x + 1) - (4y + 3) \frac{dy}{dx} = 0 $
   $ (2x + 1) = (4y + 3) \frac{dy}{dx} $
   $ \frac{2x + 1}{4y + 3} = \frac{dy}{dx} $

Therefore, $ \frac{dy}{dx} = \frac{2x + 1}{4y + 3} $.